Handout II

7/21/2019 Handout II

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Mathematics 55 - Elementary Analysis III First Semester, A.Y. 2014-2015

Summary Granario

I. Double integrals

a. Let f (x, y) and R = [a, b] × [c, d] be given. The double integral of f is the limit

R

f (x, y) dA = lim∆→0

ni=1

m j=1

f (x∗i , y∗ j )∆xi∆y j ,

where ∆ = max1≤i≤n,1≤ j≤m

x2i + y2 j , provided that the limit exists. In this case, we say that f is

integrable.

b. Partial integration:

i. b

a

f (x, y)dx - Integrate the function with respect to x with y fixed.

ii.

dc

f (x, y) dy - Integrate the function with respect to y with x fixed.

Note. The integral

ba

f (x, y) dx is a function of y and the integral

ba

f (x, y) dx is a function

of x.

Exercise. Evaluate the following integrals.

a.

10

x2y + xy2 + 2x− 3y + 1 dx

b.

21

ln(x + y) + 2x2ey dy

c.

π0

cos(x− y) + sin(xy) dx

d.

10

2x

x2 + y2 + 1 dy

c. Theorem (Fubini). Let R = [a, b] × [c, d] and let f (x, y) be continuous on R. Then

R

f (x, y) dA = ba

dc

f (x, y) dydx = dc

ba

f (x, y)dx.

Exercise. Find the volume of the solid under z = f (x, y) over the closed rectangular region R.

a. f (x, y) = ye−xy ; R = [−1, 1] × [0, 1]

b. f (x, y) = cos(3x) sin(2y) ; R = [−π, 0] × [0, π2 ]

c. f (x, y) = 1

1 + x + y ; R = [1, 3] × [1, 2]

d. A region D ⊆ R2 is said to be bounded if D ⊆ R for some closed rectangular region R (you

can enclose D with a rectangle). A Type I region D ⊆ R2 is a region of the form



D = {(x, y) ∈ R2 : a ≤ x ≤ b, g1(x) ≤ y ≤ g2(x)},

where g1, g2 are continuous functions. In this case,

D

f (x, y)dA =

ba

g2(x)g1(x)

f (x, y) dy dx.

A Type II region D ⊆ R2 is a region of the form

D = {(x, y) ∈ R2 : f 1(y) ≤ x ≤ f 2(y), c ≤ y ≤ d},

where f 1, f 2 are continuous functions. In this case,

D

f (x, y) dA =

dc

f 2(y)f 1(y)

f (x, y) dxdy.

e. Theorem. Let D ⊆ R2 be bounded and f, g be integrable functions on D. Then

i.

D

f (x, y) + g(x, y) dA =

D

f (x, y) dA +

D

g(x, y) dA

ii.

D

cf (x, y) dA = c

D

f (x, y) dA for any c ∈ R

iii.

D

f (x, y) dA =

D1

f (x, y) dA +

D2

f (x, y) dA if D = D1 ∪ D2 and the intersection of

D1 and D2 are only boundary points of D1 and D2 (their interiors do not overlap).

iv. If f (x, y) ≥ g(x, y) on D, then D

f (x, y) dA ≥ D

g(x, y) dA.



(c)

R

x2 + y2 + 4 dA, where R is the region bounded by x2 + y2 = 4, y = x2− 4, and x = 1.

(d)

R

ex2+y2+1 dA, where R is the region between the circles x2 + y2 = 4 and x2 + y2 = 8.

(e) R

x3 + y dA, where R is the region in the first quadrant bounded by the y-axis and x2 +

y2 − 6y + 6 = 0.

(f)

R

xy − y2 dA, where R is the region outside r = 1 and inside r = 2cos 2θ.

h. Parametric surfaces. A parametric surface S is a surface defined by a set of functions

S :

x = x(u, v)

y = y(u, v)

z = z(u, v)

where (u, v) ranges over a set D, the domain of parameter. The surface S is also defined by

the vector function r(u, v) = x(u, v), y(u, v), z(u, v). If u is held constant, u = u0, r(u0, v) is a

curve on S , called a constant u-curve. Similarly, we have a constant v-curve r(u, v0) if v is

fixed, v = v0.

i. Surfaces of revolution:

i. About the x-axis:

x = x

y = f (x)cos θ

z = f (x)sin θ

, where f (x) ≥ 0.

ii. About the y-axis:

x = f (y)

y = y

z = f (y)sin θ

, where f (y) ≥ 0.

iii. About the x-axis:

x = f (z)cos θ

y = f (z)sin θ

z = z

, where f (z) ≥ 0.

Exercise.

a. Find a set of parametric equations for the portion of the cylinder y2 + z2 = 5 that extends

between x = 0 and x = 1.

b. Find a set of parametric equations for the portion of the plane y − 2z = 5 that extends

between x = 0 and x = 3.

c. Find parametric equations for the surface generated by revolving the curve y = sinx about

the x-axis.

d. Find parametric equations for the surface generated by revolving the curve y− ex = 0 about

the x-axis.

e. Give a set of parametric equations for 4x2 − 3y2 + 2z2 = 6.



j. Surface area. Let S be a parametric surface defined by r(u, v) = x(u, v), y(u, v), z(u, v). The

area of S over a bounded region R is

R

∂r∂u ×

∂r

∂v

dA.

If S is the surface defined by z = f (x, y), then the area of S over R is

R

f x(x, y)2 + f y(x, y)2 + 1 dA.

Exercise. Express the area of the given surface as an iterated integral.

a. The portion of cylinder y2 + z2 = 36 above R = [0, 4] × [−6, 6].

b. The portion of sphere x2 + y2 + z2 = 16 between z = 1 and z = 3.

c. The portion of the sphere x2 + y2 + z2 = 8 inside z = x2 + y2.

d. The portion of z = x2 + y2 inside x2 + y2 = 2x.

k. A lamina L is an indealized flat object that is thin enough to be viewed as a two-dimensional

object. L is said to be homogeneous if its composition is uniform throughout. Otherwise, we

say that L is inhomogeneous. Let δ (x, y) be the density function of L. Its mass M is given

by R

δ (x, y) dA

where R is the region of the lamina L. The center of gravity C (x, y) of L is given by

x =

1

M

R

xδ (x, y

)dA

and

y = 1

M

R

yδ (x, y) dA.

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